Answer:
1. False 2.True3.True? Unsure
Explanation:
Answer : The thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole
Explanation :
First we have to calculate the heat released by the combustion.
where,
q = heat released = ?
= specific heat of calorimeter = 
= change in temperature = 
Now put all the given values in the above formula, we get:
Thus, the heat released by the combustion = 70.43 kJ
Now we have to calculate the molar enthalpy combustion.
where,
= molar enthalpy combustion = ?
q = heat released = 70.43 kJ
n = number of moles cymene = 
Therefore, the thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole
Answer:
Baking Soda, Vinegar, Ammonia
Explanation:
Answer:
A = 674.33mmHg
B = 0.385atm
Explanation:
Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.
Mathematically,
P = kT, k = P / T
P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn
A)
Data:
P1 = 799mmHg
T1 = 50°C = (50 + 273.15) = 323.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2
P2 = (P1 × T2) / T1
P2 = (799 × 273.15) / 323.15
P2 = 674.37mmHg
The final pressure is 674.37mmHg
B)
P1 = 0.470atm
T1 = 60°C = (60 + 273.15)K = 333.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2,
P2 = (P1 × T2) / T1
P2 = (0.470 × 273.15) / 333.15
P2 = 0.385atm
The final pressure is 0.385atm
HCl Acid + Sodium Hydroxide ----> Sodium Chloride + water.
<u>Explanation</u>:
- The reaction between an acid and a base is known as a neutralization reaction. The reaction of an acid with a base to give salt, water and heat is called neutralization.
- When hydrochloric acid reacts with sodium hydroxide, sodium chloride and water are produced.
HCl + NaOH → NaCl + H2O + Heat
- The heat evolved in the neutralization reaction raises the temperature of the reaction mixture.
- An electro-electrodialysis process (EED) is utilized to create HCl and NaOH from exchange NaCl. NaOH and HCl arrangements with purity higher than 99.9% are acquired. The experimental estimations of the transitions for HCl and NaOH are contrasted and values determined from the incorporation of the Nernst–Planck electro dispersion conditions.
